如何使用 awk 或 sed 获取两个单词之间的文本

Question

我有字符串列表：

./SolutionController.php core/app/Http/Controllers/Admin/SolutionController.php
./ContentController.php core/app/Http/Controllers/Frontpage/ContentController.php
./country-flag vendor/country-flag

我希望我能得到“./”符号和“空格”之间的最终值

Output：

SolutionController.php
ContentController.php
country-flag

此代码带有 bash 脚本：

#!/bin/bash
tanggal=$(date +%d-%m-%Y)
filename="./update/$tanggal/lists.md"
n=1
tanggalWaktu=$(date +"%d-%m-%Y %H:%M:%S")
mkdir -p ./logs

while read line; do
    fileName=$(awk -F'[/ ]' '{print $2}' $line) 
    echo "file -> $fileName"
done < $filename

Output：

awk: can't open file ./SolutionController.php
 source line number 

请帮我

Answer 1

使用 awk：

awk -F'[/ ]' '{print $2}' string.txt 

使用 gawk：

awk '{print gensub(/\.\/(.*) (.*)/,"\\1","g")}' string.txt

试验结果：

$ cat string.txt 
./SolutionController.php core/app/Http/Controllers/Admin/SolutionController.php
./ContentController.php core/app/Http/Controllers/Frontpage/ContentController.php
./country-flag vendor/country-flag  

$ awk -F'[/ ]' '{print $2}' string.txt 
SolutionController.php
ContentController.php
country-flag

$ awk '{print gensub(/\.\/(.*) (.*)/,"\\1","g")}' string.txt 
SolutionController.php
ContentController.php
country-flag

Answer 2

你能行的

echo "./SolutionController.php core/app/Http/Controllers/Admin/SolutionController.php" | sed -r 's/\.\/(.*) .*/\1/'

如果您将其存储在文件中。

sed -r 's/\.\/(.*) .*/\1/' strings.txt